A car has a velocity of $10 ms^{-1}$. It accelerates at $0.2 ms^{-2}$ for half a minute. Find the distance traveled during this time and the final velocity of the car. $(390 m, 16 ms^{-1})$
Difficulty: Easy
Solution:
Initial velocity = Vi = $10 ms^{-1}$
Acceleration a = 0.2 $ms^{-2}$
Time t = 0.5 min = $0.5 \times60$ = 30 s
(I) Distance S =?
(II) Final velocity Vf =?
S = $Vi \times t + \frac{1}{2} \times a \times t2$
S= $10 \times 30 + \frac{1}{2} \times 0.2 \times (30)^{2}$
S = $300 + \frac{1}{2}\times 0.2 \times 90$
S = $300 + \frac{1}{2} \times \frac{2}{10} \times$ 90
S = 300 + 90
S = 390 m
(II)Using 1st equation of motion
Vf = Vi + at
Vf = $10 + 0.2 \times30$
Vf = 10 + 6
Vf = $16 ms^{-1}$
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